# Is this problem about acid-base titration wrong?

I wouldn't post mere 'problems' here normally, but I will hopefully be starting to give chemistry education to some 10-11-12th graders and I want to make sure that it is not I that has a problem but rather the problem in the book itself. Here is the problem:

$$100~\mathrm{mL}$$ of $$\ce{H2SO4}$$ solution is titrated with $$5\cdot 10^{-2}\ \mathrm{M}$$ solution of $$\ce{NaOH}$$. Here are the information that are given in the titration graph:

Total volume of $$\ce{NaOH}$$ solution added: $$0$$ | pH: $$x$$

Total volume of $$\ce{NaOH}$$ solution added: $$400$$ | pH: $$7$$

Total volume of $$\ce{NaOH}$$ solution added: $$600$$ | pH: $$12$$

According to what has been provided above, what is the $$\mathrm{pH}$$ of the initial $$\ce{H2SO4}$$ solution ($$x$$) ? (Assume that $$\ce{H2SO4}$$ fully dissociates in water to give $$\ce{2H+}$$ and $$\ce{SO4^2-}$$ ions)

A) 1

B) 2

C) 3

D) 4

E) 5

My answer: $$-\log (2\cdot 10^{-1})$$

My steps:

1. The $$\ce{H+}$$ ions in the acid solution have been neutralized by $$400\ \mathrm{mL}$$ of $$5\cdot 10^{-2}\ \mathrm{M}\ \ce{NaOH}$$ solution. That means that the amount of $$\ce{OH-}$$ ions in $$400~\mathrm{mL}$$ of that $$\ce{NaOH}$$ solution must equal the amount of $$\ce{H+}$$ ions in the acid solution.

2. The amount of $$\ce{OH-}$$ ions in $$400~\mathrm{mL}$$ of $$5\cdot 10^{-2}~\mathrm{M}\ \ce{NaOH}$$ solution can be found by taking the product of the liter of solution and the molarity of the solution. And that gives us $$0.4\times 5\cdot 10^{-2} = 2\cdot 10^{-2}\ \mathrm{mol}\ \ce{OH-}$$ ions. (Yes, I didn't use the units, I don't have to, I explained with words what I was going to do)

3. Since we used $$2\cdot 10^{-2}\ \mathrm{mol}\ \ce{OH-}$$ ions to neutralize the acid solution, that should mean that there were $$2\cdot 10^{-2}\ \mathrm{mol}\ \ce{H+}$$ in the acid solution.

4. To find the pH of the acid solution, we need to find the concentration of $$\ce{H+}$$ ions in that solution. The volume of the solution is known, $$100~\mathrm{mL}$$. And the amount of $$ce{H+}$$ ions too is known, $$2\cdot 10^{-2}\ \mathrm{mol}$$. The concentration of $$\ce{H+}$$ ions in the acid solution should therefore be $$2\cdot 10^{-2} / 0.1 = 2\cdot 10^{-1}\ \mathrm{M}$$.

5. $$-\log[\ce{H+}] = -\log (2\cdot 10^{-1}) =$$ none of the choices above.

Your calculations are correct. I also did a second sanity check: Adding $200~\mathrm{ml}\ \ce{NaOH}$ to a neutral solution does indeed give a $\mathrm{pH}\ 12$ solution, so they also wanted you to assume that the neutral point is the equivalence point.

Which answer to select? Well, considering that:

$$-\log(2 \cdot 10^{-2}) \approx 1.699 \approx 2$$

I would probably go with $2$. Along with telling the pupils that this is a badly worded question and in their exam there will be clearly solveable ones.

• This could also be viewed as an exercise in significant figures. There is just one significant figure in $5\cdot 10^{-2}\ \mathrm{M}$ solution of $\ce{NaOH}$ and one significant figure in a pH of 7.
– MaxW
Aug 21, 2016 at 15:37
• @MaxW True, but there is only one sig fig in $2\log 2$, too.
– Jan
Aug 21, 2016 at 15:41
• that was my point. For one significant figure 1.699 rounds to 2. So it doesn't seem that the question is really "badly worded."
– MaxW
Aug 21, 2016 at 15:50
• @MaxW Hmm … When in doubt, I would prefer the single sig fig $2\log 2$ over the other single sig fig … Fell free to add your view of the topic as an additional answer, though ;)
– Jan
Aug 21, 2016 at 15:52